# is decoherence continuous?

Pardon my naivete here. In a quantum system, it seems that even a few photons from the environment can decohere the entangled particles in the system in a trillion trillionth of a second ( or faster).

But surely one decoherence is not final. As other photons interact with the system, does decoherence keep happening? It is sometimes depicted in pop physics books as like a movie, with a succesion of frames ( or successive decoherences) continually creating the classical universe. Is this clumsy metaphor in any way accurate?

To understand decoherence one must understand coherence.

In systems described by wave functions , i.e. sinusoidal distributions (waves) in a variable, coherence means that the waves , if more than one , have a known phase to each other. There is a long article in wiki on coherence.

Quantum mechanical solutions of a system of particles are wave functions, because the differential equations that describe the behavior of particles are wave equations. The only difference comes in the interpretation of the amplitude of the wave: in quantum mechanics it is a probability wave, i.e. it describes the probability of finding a particle at a particular (x,y,z) at time t.

The whole system mathematically , after solving for the potentials and boundary conditions, is described with known phases. That means a coherent system. Coherent systems may be pure, or mixed.

When there are large number of particles involved and thus many variables, the density matrix formulation allows to see clearly what parts are coherent and what parts have decohered, i.e. the phase information has been lost. In the density matrix each wavefunction describing a particle occupies nondiagonal components of the matrix according to whether the phases are known or not . If the phases are lost the nondiagonal element is zero. A completely decohered system has just diagonal elements in the density matrix. ( see the answers here for further on the density matrix).

But surely one decoherence is not final.

Decoherence when it happens is irreversible as far as the measurements go. Once the information of the phase is lost for the system, that's it.

As other photons interact with the system, does decoherence keep happening?

If they interact with the still cohered parts of the system, decoherence will keep happening until everything is diagonal. Once it is diagonal, i.e. the phase informations are lost then this is irreversible.

It is sometimes depicted in pop physics books as like a movie, with a succesion of frames ( or successive decoherences) continually creating the classical universe. Is this clumsy metaphor in any way accurate?

Well, the underlying framework of physical reality is quantum mechanical. In theory one could have one state function for the universe, including time variations and all phases known. The fact is that every measurement/interaction has finite accuracies . The system is decohered if the off diagonal elements are zero within our possibilities of measuring/ experimenting. It is the small value of h_bar and the large dimensions of macroscopic systems that very rapidly reduce the off diagonal elements to zero as far as detecting them possibilities go . But the simple picture you describe above is not good, in my opinion.

It is coherence that is necessary to build Maxwell's electromagnetic waves from individual photons, for example. That is continually going on in the macroscopic environment. So both coherence and incoherence play a role. Incoherence can be related to entropy, a classical thermodynamics concept that defines the arrow of time.

Anna's explanation is very illuminating, but here is a shorter one: Assume that we prepare a pair of particles in the entangled state

$$\exp(i\theta)\left|A_1\right\rangle\left|B_2\right\rangle + \exp(i\varphi)\left|A_2\right\rangle\left|B_1\right\rangle,$$

where subscript 1 means that the particle follows a trajectory slightly different than the particle with subscript 2. Now, we want both trajectories of particle A to fly to an experimenter Alice, and both trajectories of particle B to fly to an experimenter Bob. Please notice the simple phase factors $\exp(i\theta)$ and $\exp(i\varphi)$ .

But if we don't isolate the particles enough, i.e. we don't place the installation in high vacuum, etc., it may happen that on some path, e.g. the on the trajectory 1 toward Alice, a photon falls on the particle and is kicked away. So, we get

$$\exp(i\theta) \left|ph_{disturbed}\right\rangle\left|A_1\right\rangle\left|B_2\right\rangle + \exp(i\varphi) \left|ph_{undisturbed}\right\rangle\left|A_2\right\rangle\left|B_1\right\rangle .$$

Now we have a 3particle entanglement, and the phase factor appearing in front of each two-state product is no more a simple exponential, it is a complication. If we are aware of the undesired photon and we can eliminate it from the entanglement, (e.g. by placing mirrors to send its two forms back and undo its effect) we return our two particles to the initial entanglement. But if not, the two-particle entanglement is lost.

However, if a big number of photons interact with our entangled particles, the situation is much worse

$$\exp(i\theta) \left|ph_{X,disturbed}\right\rangle\left|ph_{Y,disturbed}\right\rangle\left|ph_{Z,disturbed}\right\rangle...\left|A_1\right\rangle\left|B_2\right\rangle + \\ \exp(i\varphi) \left|ph_{X,undisturbed}\right\rangle\left|ph_{Y,undisturbed}\right\rangle\left|ph_{Z,undisturbed}\right\rangle...\left|A_2\right\rangle\left|B_1\right\rangle .$$

See what we got instead of our initial and simple phase-factors. We call this DECOHERENCE. We can't we keep track of the photons and undo the damage.

Good luck !

• What I believe to be missing from this explanation is that the photon states have to be approximately orthogonal. This might not be the case if you do your experiment in a very small box. – lionelbrits Dec 4 '14 at 8:09
• Nice try with $\LaTeX$, @Sofia . I've improved it a bit more, hope you don't mind. – Ruslan Dec 4 '14 at 9:57